Answer
$$y' =1+\frac{1 }{y^2}$$
Work Step by Step
Given $$y=x+\tan^{-1}y$$
Differentiate both sides
\begin{align*}
y'&=1+\frac{y'}{1+y^2}\\
y'\left(1-\frac{1}{1+y^2}\right)&=1\\
y'\left( \frac{ y^2}{1+y^2}\right)&=1\\
y'&=\frac{1+y^2}{y^2}\\
&=1+\frac{1 }{y^2}
\end{align*}